ANSWER
Let DE and DF be the perpendiculars from D on AB and AC respectively.
In △s BDE and CDF, DE=DF (Given)
∠BED=∠CFD=90
∘
BD=DC (∵ D is the mid-point of BC)
∴ △BDE≅△CDF (RHS)
⇒ ∠B=∠C (cpct)
⇒ AC=AB (Sides opp. equal ∠s are equal)
⇒ △ABC is isosceles
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Answers & Comments
ANSWER
Let DE and DF be the perpendiculars from D on AB and AC respectively.
In △s BDE and CDF, DE=DF (Given)
∠BED=∠CFD=90
∘
BD=DC (∵ D is the mid-point of BC)
∴ △BDE≅△CDF (RHS)
⇒ ∠B=∠C (cpct)
⇒ AC=AB (Sides opp. equal ∠s are equal)
⇒ △ABC is isosceles